Electrical Installation Calculations: Advanced

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4 Electrical Installation Calculations: Advanced FOR TECHNICAL CERTIFICATE AND NVQ LEVEL 3 SEVENTH EDITION A. J. WATKINS CHRIS KITCHER AMSTERDAM BOSTON HEIDELBERG LONDON NEW YORK OXFORD PARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO Newnes is an imprint of Elsevier

5 Newnes is an imprint of Elsevier Linacre House, Jordan Hill, Oxford OX2 8DP 30 Corporate Drive, Burlington MA First edition 1957 Fifth edition 1999 Reprinted 2001, 2002, 2003, 2004 Sixth edition 2006 Seventh edition 2009 Copyright 2009, Chris Kitcher and Russell K. Parton. All rights reserved The right of Chris Kitcher and Russell K. Parton to be identified as the authors of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988 No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) ; fax (+44) (0) ; permissions@elsevier.com. Alternatively you can submit your request online by visiting the Elsevier website at and selecting Obtaining permission to use Elsevier material Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences, in particular, independent verification of diagnoses and drug dosages should be made British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging-in-Publication Data A catalog record for this book is available from the Library of Congress ISBN For information on all Newnes publications visit our website at Printed and bound in Italy

6 Malestrom Contents Preface vii Use of calculators 1 Simple transposition of formulae 3 SI units 5 Conductor colour identification 8 Alternating current circuit calculations 9 Impedance 9 Inductive reactance 13 Capacitive reactance 16 Impedance in series circuits 20 Impedance triangles and power triangles 29 a.c. waveform and phasor representation 43 Alternating e.m.f. and current 43 Phasors 48 Parallel circuits involving resistance, inductance and capacitance 56 Power factor improvement 64 Calculation without p.f. correction 66 Calculation with p.f. correction 66 Three-phase circuit calculations 70 Star-connected motors 70 Delta-connected motors (mesh) 72 Resistance and inductance in three-phase circuits 74 Three-phase circuits 78 Three-phase power 80 Voltage drop in three-phase circuits 85

7 Contents Voltmeters and ammeters: changing the use and extending the range 103 Voltmeters 103 Ammeters 105 Alternating current motors 109 Application of diversity factors 116 Cable selection 122 Earthing conductor calculation 122 Voltage drop and cable selection calculations 126 Earth leakage protection calculations 140 Lighting calculations 147 Units and quantities 147 Inverse square law 147 Cosine law 149 Mechanics 154 Moment of force 154 Torque 154 Power 155 Surface speed, pulley diameter and speed ratios 157 Miscellaneous examples 162 d.c. generators 162 d.c. motors 163 Alternators and synchronous motors 164 Induction motors 165 Insulation resistance 166 Formulae 170 Glossary 176 Answers to exercises 180 General questions 192 Additional questions 200 vi

8 Preface Mathematics forms the essential foundation of electrical installation work. Without applying mathematical functions we would be unable to work out the size of a room which needs lighting or heating, the size and/or the number of the lights or heaters themselves, the number and/or the strength of the fixings required, or the size of the cables supplying them. We would be unable to accurately establish the rating of the fuse or circuit breaker needed to protect the circuits, or predict the necessary test results when testing the installation. Like it or not you will need to be able to carry out mathematics if you want to be an efficient and skilled electrician. This book will show you how to perform the maths you will need to be a proficient electrician. It concentrates on the electronic calculator methods you would use in class and in the workplace. The book does not require you to have a deep understanding of how the mathematical calculations are performed; you are taken through each topic step by step, then you are given the opportunity yourself to carry out exercises at the end of each chapter. Throughout the book useful references are made to the 17th edition BS 7671: 2008 Requirements for Electrical Regulations and the IEE On-Site Guide. Electrical Installation Calculations : Advanced originally written by A. J. Watkins and R. K. Parton has been the preferred book for many students looking to improve their mathematical understanding of the subject for many years. This edition has been newly updated to the 17th edition IEE Wiring Regulations, not only to include modern methods, but also to cover all aspects of the new City and Guilds 2330 Certificate in Electrotechnical Technology. This second volume includes advanced calculations, in particular those involving cable selection. It will be of considerable use to those already involved in electrical installation work, as well as being invaluable to students studying the City and Guilds , 202, 203, and 205 but in particular the 302, and 303. It will be particularly useful to those studying for the City and Guilds Inspection and test Design and verification, as well as the th edition wiring regulation exams. The book also contains a variety of questions and answers to help students with the City & Guilds 2330 GOLA exams. Chris Kitcher

9 Use of calculators Throughout the Basic and Advanced books, the use of a calculator is encouraged. Your calculator is a tool, and like any tool practice is required to perfect its use. A scientific calculator will be required, and although they differ in the way the functions are carried out, the end result is the same. The examples are given using a Casio fx-83ms. The figures printed on the button is the function performed when the button is pressed. To use the function in small letters above any button the shift button must be used. Practice is important Syntax error x 2 x 3 3 x 1 Appears when the figures are entered in the wrong order. Multiplies a number by itself, i.e. 6 6 = 36. On the calculator this would be 6x 2 = 36. When a number is multiplied by itself it is said to be squared. Multiplies a number by itself and then the total by itself again, i.e. when we enter 4 on calculator x 3 = 64. When a number is multiplied in this way it is said to be cubed. Gives the number which achieves the total by being multiplied by itself, i.e. 36 = 6. This is said to be the square root of a number and is the opposite of squared. Gives you the number which when multiplied by itself three times will be the total = 4 this is said to be the cube root. 1 Divides 1 by a number, i.e. 4 = This is the reciprocal button and is useful in this book for finding the resistance of resistors in parallel and capacitors in series. 1

10 Electrical Installation Calculations: Advanced EXP The powers of 10 function, i.e = 25 EXP 10 3 = Enter into calculator 25 EXP 3 = (Do not enter the or the number 10.) If a calculation shows 10 3, i.e enter 25 EXP 3 = (0.025) (when using EXP if a minus is required use the button ( )) Brackets These should be used to carry out a calculation within a calculation. Example calculation: 32 ( ) = Enter into calculator 32 ( )= Remember, Practice makes perfect! 2

11 Simple transposition of formulae To find an unknown value: The subject must be on the top line and must be on its own. The answer will always be on the top line. To get the subject on its own, values must be moved. Any value that moves across the = sign must move from above the line to below line or from below the line to above the line. Example = = 2? Transpose to find? = 6 Example 2 2 6? = 4 Step = 4? Step =? 3

12 Electrical Installation Calculations: Advanced Answer = 3 Example = 3 20? Step 1: move 3 20 away from the unknown value, as the known values move across the = sign they must move to the bottom of the equation =? Step 2: Carry out the calculation Therefore or = = = = =

13 SI units In Europe and the UK, the units for measuring different properties are known as SI units. SI stands for Système Internationale. All units are derived from seven base units. Base quantity Base unit Symbol Time Second s Electrical current Ampere A Length Metre m Mass Kilogram kg Temperature Kelvin K Luminous intensity Candela cd Amount of substance Mole mol SI-derived units Derived quantity Name Symbol Frequency Hertz Hz Force Newton N Energy, work, quantity of heat Joule J Electric charge, quantity of Coulomb C electricity Power Watt W Potential difference, electromotive force Volt VorU 5

14 Electrical Installation Calculations: Advanced Capacitance Farad F Electrical resistance Ohm Magnetic flux Weber Wb Magnetic flux density Tesla T Inductance Henry H Luminous flux Lumen cd Area Square metre m 2 Volume Cubic metre m 3 Velocity, speed Metre per second m/s Mass density Kilogram per kg/m 3 cubic metre Luminance Candela per square metre cd/m 2 SI-Unit prefixes Name Multiplier Prefix Power of 10 Tera T Giga G Mega M Kilo 1000 k Unit 1 Milli m Micro Nano Pico Examples ma km v GW kw Milliamp = one thousandth of an ampere Kilometre = one thousand metres Microvolt = one millionth of a volt Gigawatt = one thousand million watts Kilowatt = one thousand watts 6

15 SI units Calculator example 1 kilometre is 1 metre 10 3 Enter into calculator 1 EXP 3 = (1000) metres 1000 metres is 1 kilometre 10 3 Enter into calculator 1000 EXP 3 = (1) kilometre 1 microvolt is 1 volt 10 6 Enter into calculator 1 EXP 6=(1 06 or ) volts (note sixth decimal place). 7

16 Conductor colour identification Old colour New colour Marking Phase 1 of a.c. Red Brown L1 Phase 2 of a.c. Yellow Black L2 Phase 3 of a.c. Blue Grey L3 Neutral of a.c. Black Blue N Note: great care must be taken when working on installations containing old and new colours. 8

17 Alternating current circuit calculations Impedance In d.c. circuits, the current is limited by resistance. In a.c. circuits, the current is limited by impedance (Z). Resistance and impedance are measured in ohms. For this calculation, Ohm s law is used and Z is substituted for R (Figure 1). U Z = I or voltage (U) impedance (ohms) = current (amperes) Example 1 The voltage applied to a circuit with an impedance of 6, is 200 volts. Calculate the current in the circuit. U Z = I = A Example 2 The current in a 230 V single phase motor is 7.6 A. Calculate the impedance of the circuit. 9

18 Electrical Installation Calculations: Advanced V I Z Figure 1 Volt drop U I = Z = Example 3 A discharge lamp has an impedance of 265 and the current drawn by the lamp is 0.4 A. Calculate the voltage. Z I = = U 106 volts Example 4 The current through an impedance of 32 is 8 A. Calculate the voltage drop across the impedance. U = I Z = 8 32 = 256 V 10

19 Alternating current circuit calculations Example 5 The current through an electric motor is 6.8 A at 230 V. Calculate the impedance of the motor. U = I Z (Transpose for Z) Z = U I = = Example 6 An a.c. coil has an impedance of 430. Calculate the voltage if the coil draws a current of 0.93 A. U U = = = = I Z I Z V Exercise 1 1. Complete the following table: Volts (a.c.) Current (A) Impedance () Complete the following table: Current (A) Volts (a.c.) Impedance ()

20 Electrical Installation Calculations: Advanced 3. Complete the following table: Impedance () Volts (a.c.) Current (A) A mercury vapour lamp takes 2.34 A when the mains voltage is 237 V Calculate the impedance of the lamp circuit. 5. An inductor has an impedance of 365. How much current will flow when it is connected to a 400 V a.c. supply? 6. A coil of wire passes a current of 55 A when connected to a 120 V d.c. supply but only 24.5 A when connected to a 110 V a.c. supply. Calculate (a) the resistance of the coil, (b) its impedance. 7. Tests to measure the impedance of an earth fault loop were made in accordance with BS 7671 and the results for five different installations are given below. For each case, calculate the value of the loop impedance. Test voltage, a.c. (V) Current (A) (a) (b) (c) (d) (e) The choke in a certain fluorescent-luminaire fitting causes a voltage drop of 150 V when the current through it is 1.78 A. Calculate the impedance of the choke. 9. Complete the following table: Volts (a.c.) Current (A) Impedance () Plot a graph showing the relationship between current and voltage. From the graph, state the value of the current when the voltage is 240 V. 10. The alternating voltage applied to a circuit is 230 V and the current flowing is A. The impedance of the circuit is (a) 5.4 (b) 1840 (c) 3.5 (d) An alternating current of 2.4 A flowing in a circuit of impedance 0.18 produces a voltage drop of (a) V (b) 13.3 V (c) V (d) 4.32 V 12

21 Alternating current circuit calculations 12. When an alternating e.m.f. of 150 V is applied to a circuit of impedance 265, the current is (a) A (b) 1.77 A (c) 5.66 A (d) A Inductive reactance When an a.c. current is passed through a conductor, a magnetic field is created around the conductor. If the conductor is wound into a coil the magnetic field is increased. Where there are significant magnetic fields in a circuit there is opposition to the flow of current, this opposition is called inductive reactance. The opposition caused by inductive reactance is in addition to the opposition caused by the resistance caused by the wires. In this section, we will assume that the resistance of the circuits is so low that it may be ignored and that the only opposition to the flow of current is that caused by the inductive reactance. The formula for inductive reactance is X L = 2fL (answer in ohms). Where L is the inductance of the circuit or coil of wire and is stated in henrys (H), f is the frequency of the supply in hertz (Hz) (Figure 2). H I Figure 2 13

22 Electrical Installation Calculations: Advanced Example 1 Calculate the inductive reactance of a coil which has an inductance of 0.03 henrys when connected to a 50 Hz supply. X L = 2fL = = 9.42 Example 2 Calculate the inductive reactance of the coil in Example 1 when connected to a 60 Hz supply. X L = 2fL = = It can be seen from this calculation that if the frequency increases the inductive reactance will also increase. Example 3 An inductor is required to cause a voltage drop of 180 volts when a current of 1.5 A is passed through it at a frequency of 50 Hz. Calculate the value of the inductor: U L = I X L (this is Ohm s law with inductive reactance instead of resistance) 14

23 Alternating current circuit calculations Transposed U I = X L Transpose = 120 X L = 2fL 120 = L 120 ( ) = H On calculator enter 120 (2 50) = (answer H) Exercise 2 1. Calculate the inductive reactance of a coil having an inductance of H when a 50 Hz current flows in it. 2. A coil is required to have an inductive reactance of 150 ona50hz supply. Determine its inductance. 3. Complete the following table: Inductance (H) Frequency (Hz) Reactance () A coil of negligible resistance causes a voltage drop of 98 V when the current through it is 2.4 A at 50 Hz. Calculate (a) the reactance of the coil, (b) its inductance. 5. A reactor allows a current of 15 A to flow from a 230 V 50 Hz supply. Determine the current which will flow at the same voltage if the frequency changes to (a) 45 Hz, (b) 55 Hz. Ignore the resistance. 15

24 Electrical Installation Calculations: Advanced 6. Calculate the inductive reactance of coils having the following values of inductance when the supply frequency is 50 Hz. (a) H (b) H (c) 0.45 mh (d) 350 H (e) H 7. Determine the inductances of the coils which will have the following reactance to a 50 Hz supply: (a) 300 (b) 25 (c) 14.5 (d) 125 (e) 5 8. The inductance of a coil can be varied from 0.15 H to 0.06 H. Plot a graph to show how the inductive reactance varies with changes in inductance. Assume a constant frequency of 50 Hz. 9. A reactor has a constant inductance of 0.5 H and it is connected to a supply of constant voltage 100 V but whose frequency varies from 25 to 50 Hz. Plot a graph to show how the current through the coil changes according to the frequency. Ignore the resistance of the coil. 10. Calculate the voltage drop across a 0.24 H inductor of negligible resistance when it carries 5.5 A at 48 Hz. 11. An inductor of H is connected to an a.c. supply at 50 Hz. Its inductive reactance is (a) 39.3 (b) 0.79 (c) (d) The value in henrys of an inductor which has an inductive reactance of 500 when connected in an a.c. circuit at frequency 450 Hz is (a) 1.77 H (c) H (b) H (d) H Capacitive reactance When a capacitor is connected to an a.c. supply, the current flow is limited by the reactance of the capacitor (X C ). Formula for capacitive reactance X C = 106 2fC where C is the capacitance of the capacitor measured in microfarads (F) and f is the frequency of the supply in hertz (Hz). (It should be 16

25 Alternating current circuit calculations C I Hz Figure 3 noted that d.c. current will not flow with a capacitor in the circuit it will simply charge and then stop.) (Figure 3) Example 1 Calculate the reactance of a 70 F capacitor to a 50 Hz supply: X C = 106 2fC = Enter on calculator EXP 6 ( ) = (answer 45.47). Example 2 A power factor improvement capacitor is required to take a current of 7.5 A from a 230 volt 50 Hz supply. Determine the value of the capacitor. For this calculation, Ohm s law is used and R is substituted by X C. Step 1 U C = I X C 230 = 7.5 X C 17

26 Electrical Installation Calculations: Advanced Transpose for X C = X C = 30.6 Step2tofindC X C = 106 2fC Transpose C = f X C C = 10 6 ( ) = 104 answer in microfarads (F) (Note simply change places of X C and C) Enter on calculator EXP 6 ( ) or EXP 6 ( ) Exercise 3 1. Determine the reactance of each of the following capacitors to a 50 Hz supply. (Values are all in microfarads.) (a) 60 (d) 150 (g) 250 (j) 75 (b) 25 (e) 8 (h) 95 (c) 40 (f) 12 (i) Calculate the value of capacitors which have the following reactances at 50 Hz. (Values are all in ohms). (a) 240 (d) 4.5 (g) 45 (j) 72 (b) 75 (e) 36 (h) 400 (c) 12 (f) 16 (i) 30 18

27 Alternating current circuit calculations 3. Calculate the value of a capacitor which will take a current of 25 A from a 230 V 50 Hz supply. 4. A capacitor in a certain circuit is passing a current of 0.2 A and the voltage drop across it is 100 V. Determine its value in microfarads. The frequency is 50 Hz. 5. A 20 F capacitor is connected to an alternator whose output voltage remains constant at 150 V but whose frequency can be varied from 25 to 60 Hz. Draw a graph to show the variation in current through the capacitor as the frequency changes over this range. 6. Calculate the voltage drop across a 5 F capacitor when a current of 0.25 A at 50 Hz flows through it. 7. In order to improve the power factor of a certain installation, a capacitor which will take 15 A from the 230 V supply is required. The frequency is 50 Hz. Calculate the value of the capacitor. 8. In one type of twin-tube fluorescent fitting, a capacitor is connected in serieswithoneofthetubes.ifthevalueofthecapacitoris7f, the current through it is 0.8 A, and the supply is at 50 Hz, determine the voltage across the capacitor. 9. A machine designed to work on a frequency of 60 Hz has a power-factor-improvement capacitor which takes 12 A from a 110 V supply. Calculate the current the capacitor will take from the 110 V 50 Hz supply. 10. A capacitor takes a current of 16 A from a 400 V supply at 50 Hz. What current will it take if the voltage falls to 380 V at the same frequency? 11. A 22 F capacitor is connected in an a.c. circuit at 50 Hz. Its reactance is (a) (c) (b) 6912 (d) The value in microfarads of a capacitor which has a capacitive reactance of 100 when connected to a circuit at 50 Hz is (a) 31.8 F (c) F (b) 318 F (d) F 19

28 Electrical Installation Calculations: Advanced H R I Hz Figure 4 Impedance in series circuits When resistance (R) is in a circuit with reactance (X L or X C ), the combined effect is called Impedance (Z), this is measured in ohms. For series circuits, the calculation for impedance (Z) is Z 2 = R 2 + X 2 or Z = R 2 + X 2 In this calculation X is for X C or X L. Where the circuit contains inductive reactance (X C ) and capacitive reactance (X L ). X = X C X L or X L X C X will be the largest reactance minus the smallest reactance. An inductor coil will always possess both inductance (the magnetic part of the circuit) and resistance (the resistance of the wire), together they produce impedance. Although inductance and impedance cannot be physically separated, it is convenient for the purpose of calculation to show them separately in a circuit diagram (Figure 4). Example 1 A coil has a resistance of 6 and an inductance of 0.09 H. Calculate its impedance to a 50 Hz supply. Step 1 Inductive reactance X L =2fL 20

29 Alternating current circuit calculations 2 f = (Note: a common error is to add the resistance and inductance treating it as a d.c. circuit.) Step 2 Z 2 = R 2 + X 2 L or Z = Z = R 2 + X = Enter into calculator 6X X 2 = = (answer 28.9 ). Example 2 A coil passes a current of 23 A when connected to a 230 V d.c. supply, but only 8 A when connected to a 230 V a.c. supply. When connected to a d.c. circuit the coil s resistance is only that of the wire in the coil, this can be calculated using Ohm s law. On d.c. U = I R U I = R = 10 (resistance) On an a.c. circuit, reactance will be produced, as this is an inductive circuit it will be inductive reactance (X L ). The combined effect of the resistance and reactance of the coil is the impedance (Z). Step 1 On a.c.u = I Z 230 = 8 Z 21

30 Electrical Installation Calculations: Advanced Transpose = impedance (Z). Step 2 To find the inductance of the coil. Z 2 = R 2 + X 2 L X 2 L = Z2 R 2 X L = X L = Enter on calculator 28.7X = = (answer ) Step 3 X L = 2fL = L Transpose ( ) = L = H Enter on calculator ( ) = (answer) Example 3 A70 resistor is wired in series with a capacitor of an unknown value to a 230 volt 50 Hz supply. Calculate the value of the capacitor in microfarads if a current of 1.3 A flows. First find impedance of circuit (Z) 22

31 Alternating current circuit calculations C μf? 70 Ω 1.3 A 230 V 50 Hz Figure 5 Step 1(Figure 5) U = I Z 230 = 1.3 Z Z = Z = Step 2 Next find capacitive reactance X C Z = = R 2 + X 2 C XC 2 Transpose for X C X C = X C = Now find capacitance Step 3 X C = 106 2fC 23

32 Electrical Installation Calculations: Advanced 0.09H 70 μf 18Ω 230 V 50 Hz Figure 6 Transpose for C C = 106 2f X L C = F is the capacitor value On calculator enter EXP 6 ( ) = (answer) Example 4 A coil of inductance of 0.09 H and a resistance of 18 is wired in series with a 70 F capacitor to a 230 volt 50 Hz supply. Calculate the current which flows and the voltage drop across the capacitor (Figure 6). Step 1 Calculate inductive and capacitive reactance. Inductive reactance 24 X L = 2fL = = 28.27

33 Alternating current circuit calculations Capacitive reactance X C = 106 2fC = = Enter on calculator EXP 6 ( ) = (answer) Step 2 Find the actual reactance for circuit which is the largest reactance minus the smallest reactance For this circuit X = X C X L = = (this is X C as the capacitive reactance is larger than the inductive reactance) Step 3 Calculate the impedance for the circuit (Z) Impedance Z is found Z 2 = R 2 + X 2 Z 2 = Z = Enter on calculator 18X X 2 = = (answer) Z = Step 4 Calculate current (I) U = I Z 230 = I

34 Electrical Installation Calculations: Advanced V r V c Figure 7 V s 230 Transpose for I = 9.24 A As this current is common to the whole circuit, the voltage across the capacitor and the inductor can be calculated. If a phasor is required the current is the reference conductor (Figure 7). Voltage across capacitor U C = I X C = = 420 volts Voltage across inductor U I = I X L = = volts (Note both voltages are higher than the 230 V supply. This often happens in a.c. circuits. The voltages do not add up as in d.c. circuits.) 26

35 Alternating current circuit calculations Exercise 4 1. Complete the following table: R R Complete the following table: X X A coil of wire has resistance of 8 and inductance of 0.04 H. It is connected to a supply of 100 V at 50 Hz. Calculate the current which flows. 4. An inductor of inductance H and resistance 12 is connected to a 230 V supply at 50 Hz. Calculate the current which flows. 5. Complete the following table: R () X () Z () A capacitor of 16 F and a resistor of 120 are connected in series. Calculate the impedance of the circuit. 7. A resistor of 200 and a capacitor of unknown value are connected to a 230 V supply at 50 Hz and a current of 0.85 A flows. Calculate the value of the capacitor in microfarads. 8. When a certain coil is connected to a 110 V d.c. supply, a current of 6.5 A flows. When the coil is connected to a 110 V 50 Hz a.c. supply, only 1.5 A flows. Calculate (a) the resistance of the coil, (b) its impedance, and (c) its reactance. 9. The inductor connected in series with a mercury vapour lamp has resistance of 2.4 and impedance of 41. Calculate the inductance of the inductor and the voltage drop across it when the total current in the circuit is 2.8 A. (Assume the supply frequency is 50 Hz.) 10. An inductor takes 8 A when connected to a d.c. supply at 230 V. If the inductor is connected to an a.c. supply at 230 V 50 Hz, the current is 4.8 A. Calculate (a) the resistance, (b) the inductance, and (c) the impedance of the inductor. 11. What is the function of an inductor in an alternating-current circuit? When a d.c. supply at 230 V is applied to the ends of a certain inductor coil, the current in the coil is 20 A. If an a.c. supply at 230 V 50 Hz is applied to the coil, the current in the coil is A. 27

36 Electrical Installation Calculations: Advanced Calculate the impedance, reactance, inductance, and resistance of the coil. What would be the general effect on the current if the frequency of the a.c. supply were increased? 12. A coil having constant inductance of 0.12 H and resistance of 18 is connected to an alternator which delivers 100 V a.c. at frequencies ranging from 28 to 55 Hz. Calculate the impedance of the coil when the frequency is 30, 35, 40, 45 and 50 Hz and plot a graph showing how the current through the coil varies according to the frequency. 13. The inductor in a discharge lighting circuit causes a voltage drop of 120 V when the current through it is 2.6 A. Determine the size in microfarads of a capacitor which will produce the same voltage drop at the same current value. (Neglect the resistance of the inductor. Assume the supply frequency is 50 Hz.) 14. A circuit is made up of an inductor, a resistor and a capacitor all wired in series. When the circuit is connected to a 50 Hz a.c. supply, a current of 2.2 A flows. A voltmeter connected to each of the components in turn indicates 220 V across the inductor, 200 V across the resistor, and 180 V across the capacitor. Calculate the inductance of the inductor and the capacitance of the capacitor. At what frequency would these two components have the same reactance value? (Neglect the resistance of the inductor.) 15. What are meant by the following terms used in connection with alternating current: resistance, impedance and reactance? A voltage of 230 V, at a frequency of 50 Hz, is applied to the ends of a circuit containing a resistor of 5, an inductor of 0.02 H, and a capacitor of 150 F, all in series. Calculate the current in the circuit. 16. A coil of resistance 20 and inductance 0.08 H is connected to a supply at 240 V 50 Hz. Calculate (a) the current in the circuit, (b) the value of a capacitor to be put in series with the coil so that the current shall be 12 A (CGLI) (Figure 8) X L = 24Ω X C = 20Ω 2 A Figure 8 R = 3Ω U ~ 17. For the circuit shown in Figure 8, the voltage V is (a) 94 V (b) 14 V (c) 10 V (d) V 28

37 Alternating current circuit calculations 18. An inductor has inductance 0.12 H and resistance 100. When it is connected to a 100 V supply at 150 Hz, the current through it is (a) 1.51 A (b) 0.47 A (c) 0.66 A (d) A Impedance triangles and power triangles For a right-angled triangle (Figure 9), the theorem of Pythagoras states that a 2 = b 2 + c 2 As the relationship between impedance, resistance and reactance in a series circuit is given by an equation of a similar form, Z 2 = R 2 + X 2, conditions in such circuits can conveniently be represented by right-angled triangles. In Figure 10, Z 2 = R 2 + X 2 where X = X L (Figure 10(a)) or X C (Figure 10(b)) and = the phase angle of the circuit sin = X Z cos = R Z and tan = X R cos is the power factor of the circuit (Figure 10). A right-angled triangle is also used to represent the apparent power in a circuit and its active and reactive components (Figure 11). AB is the product of voltage and current in the circuit (VA). AC is the true power the working component (W). BC is the reactive or wattless component (VAr). a b c Figure 9 29

38 Electrical Installation Calculations: Advanced Impedance Inductive Z reactance X L f Resistance R (a) Inductive reactance Figure 10 Resistance R f Capacitive reactance Impedance X C Z (b) Capacitive reactance A f Figure 11 VA W B A VAr (leading) C f W VA C VAr (lagging) B VAr VA = sin VAr = VA sin W VA W = cos = VA cos and cos ϕ is the power factor (p.f.). In power circuits, the following multiples of units are used: kva kw and kvar. Example 1 Find Z in Figure

39 Alternating current circuit calculations Z X L = 78 Ω R = 56 Ω Figure 12 Z 2 = R 2 + X 2 L = = = 9220 Z = 9220 = = 96 (correct to three significant figures) Example 2 Find X C in Figure 13. R = 67.6 Ω Z = 125 Ω X C Figure 13 31

40 Electrical Installation Calculations: Advanced Z 2 = R 2 + X 2 C = X 2 C X 2 C = = = X C = = 105 Alternatively, Z 2 = R 2 + X 2 C = X 2 C X 2 C = = ( )( ) = = X C = = 105 Example 3 Find in Figure 14. Z X L = 15 Ω f Figure 14 R = 20 Ω 32

41 Alternating current circuit calculations tan = X L R = = 0.75 = Example 4 Find X C in Figure 15. X C/Z = sin X C 90 = sin 48 = X C = = 66.9(to three significant figures) Example 5 Find the kva and kvar in Figure 16. R f = 48 Z = 90 Ω X C Figure 15 33

42 Electrical Installation Calculations: Advanced kva f = kw Figure 16 kvar kw kva = cos 15 kva = cos 42 = KVA 15 = kva = = 20.2 kvar kw = tan kvar 15 = tan 42 = kvar = = 13.5 Example 6 A coil of 0.2 H inductance and negligible resistance is connected in series with a 50 resistor to the 230 V 50 Hz mains (Figure 17). Calculate (a) the current which L 0.2 H R 50 Ω U 230V 50Hz Figure 17 34

43 Alternating current circuit calculations Z X L f R Impedance triangle Figure 18 flows, (b) the power factor, (c) the phase angle between the current and the applied voltage. Coil reactance X L = 2fL = = = 62.8 To find the impedance (Figure 18) Z 2 = R 2 + X 2 L = = = 6444 Z = 6444 = (a) To find the current, U = I Z 230 = I I = = 2.86 A 35

44 Electrical Installation Calculations: Advanced (b) Power factor = cos = R Z = = lag (c) The phase angle is the angle whose cosine is 0.623, = Exercise 5 1. Find Z in Figure 19 Z X L = 40 Ω R = 30 Ω Figure Find Z in Figure 20 R = 25 Ω Z X C = 31.4 Ω Figure

45 Alternating current circuit calculations 3. Find R in Figure 21 Z = 130 Ω X L = 120 Ω R Figure Find X C in Figure 22 R = 135 Ω Z = 240 Ω X C Figure Find R in Figure 23 Z = 60.5 Ω X L = 39 Ω R Figure Find Z in Figure 24 R = 175 Ω Z X C = 150 Ω Figure Find R in Figure 25 Z = 31.3 Ω X L = Ω R Figure

46 Electrical Installation Calculations: Advanced 8. Find X L in Figure 26 Z = Ω X L R = Ω Figure Find Z in Figure 27 Z X C = 354 Ω Figure Find X L in Figure 28 Z = 753 Ω X L R = 50 Ω Figure Find R in Figure 29 R Z = 2620 Ω X C = 2600 Ω Figure Consider the answers to questions 9 to 11 and then write down the approximate impedance of a coil which has resistance 32 and reactance

47 Alternating current circuit calculations 13. Complete the following table: Angle sin cos tan 14. Complete the following table: Angle sin cos tan 15. Complete the following table: Angle sin cos tan Complete the following table: Angle sin cos tan Find R and X L in Figure 30 Z = 29.2 Ω X L f = R Figure Find R and X C in Figure 31 R f = Z = 7.29 Ω X C Figure

48 Electrical Installation Calculations: Advanced 19. Find in Figure 32 X L = 18.4 Ω f R = 29.7 Ω Figure Calculate Z and X L in Figure 33 Z X L f = 59 6 R = 46.7 Ω Figure Find W and VAr in Figure 34 W f = VA = 250 VAr Figure Find and X L in Figure 35 Z = 238 Ω X L f R = 200 Ω Figure Find in Figure 36 kw = 4.9 f kvar kva = 5.6 Figure

49 Alternating current circuit calculations 24. Calculate R in Figure 37 R f = 78 5 X C = 314 Ω Figure Find OX in Figure 38 O X OY = 74.6 Y Figure Find OX in Figure 39 O OY = 50 X Y Figure Complete the following table then plot a graph of power factor (cos )toa base of phase angle (): Phase angle Power factor cos A coil has inductance 0.18 H and resistance 35. It is connected to a 100 V 50 Hz supply. Calculate (a) the impedance of the coil, (b) the current which flows, (c) the power factor, (d) the power absorbed by the coil. 41

50 Electrical Installation Calculations: Advanced 29. Define the term power factor and state how it affects cable size. An inductor of resistance 8 and of inductance H is connected to an alternating-current supply at 230 V, single-phase, 50 Hz. Calculate (a) the current from the supply, (b) the power in the circuit, (c) the power factor. 30. A single-phase alternating-current supply at 230 V 50 Hz is applied to a series circuit consisting of an inductive coil of negligible resistance and a non-inductive resistance coil of 15. When a voltmeter is applied to the ends of each coil in turn, the potential differences are found to be V across the inductive coil, 203 V across the resistance. Calculate (a) the impedance of the circuit, (b) the inductance of the coil, (c) the current in the circuit, and (d) the power factor. (CGLI) 31. On what factors do the resistance, reactance and impedance of an alternating-current circuit depend, and how are these quantities related? The current in a single-phase circuit lags behind the voltage by 60.The power in the circuit is 3600 W and the voltage is 240 V. Calculate the value in ohms of the resistance, the reactance and the impedance. (CGLI) 42

51 a.c. waveform and phasor representation Alternating E.M.F. and current The value and direction of the e.m.f. induced in a conductor rotating at constant speed in a uniform magnetic field, Figure 40(a) vary according to the position of the conductor. The e.m.f. can be represented by the displacement QP of the point P above the axis XOX, Figure 40(b). OP is a line which is rotating about the point O at the same speed as the conductor is rotating in the magnetic field. The length of OP represents the maximum value of the induced voltage. OP is called a phasor. A graph, Figure 40(c), of the displacement of the point P plotted against the angle (the angle through which the conductor has moved from the position of zero induced e.m.f.) is called a sine wave, since the PQ is proportional to the sine angle. One complete revolution of OP is called a cycle. Example 1 An alternating voltage has a maximum value of 200 V. Assuming that it is sinusoidal in nature (i.e. it varies according to a sine wave), plot a graph to show the variations in this voltage over a complete cycle. Method (Figure 41) Choose a reasonable scale for OP; for instance, 10mm=100V. Draw a circle of radius 20 mm at the left-hand side of a piece of graph paper to represent the rotation of OP. One complete revolution of OP sweeps out 360. Divide the circle into any number of equal portions, say 12. Each portion will then cover 30. Construct the axes of the graph, drawing the horizontal axis OX (the x-axis) on a line through the centre of the circle. This x-axis should now be marked off in steps of 30 up to 360. If desired, perpendicular lines can be drawn through these points. Such lines are called ordinates. 43

52 Electrical Installation Calculations: Advanced N 1 cycle P + q X O q Q X e.m.f. S q (degrees) (a) (b) (c) Figure 40 Figure 41 The points on the graph are obtained by projecting from the various positions of P to the coordinate corresponding to the angle at that position. Remember that when =0 and 180 the generated e.m.f. is zero, and when =90 and 270 the generated e.m.f. has its maximum value. Example 2 Two alternating voltages act in a circuit. One (A) has an r.m.s. value of 90 V and the other (B) has an r.m.s. value of 40 V, and A leads B by 80. Assuming that both voltages are sinusoidal, plot graphs to show their variations over a complete cycle. By adding their instantaneous values together, derive a graph of the resultant voltage. Give the r.m.s. value of this resultant. First find the maximum values of the voltages given: 44 U r.m.s = U max 90 = U max U max = = 127 V

53 a.c. waveform and phasor representation Y + Resultant A B O A O Figure 42 Y Similarly, if U r.m.s. = 40 U max = = 56.6 V Choose a suitable scale, say 20 mm = 100 V. Draw two circles with the same centre, one having a radius of 25.4 mm (127 V), the other a radius of mm (56.6 V). Draw phasors to represent the voltages: OA horizontal and OB, which represents the lower voltage, lagging 80 behind OA (anticlockwise rotation is always used) see Figure 42. Mark off the circumference of the larger circle in steps of 30, using OA as the reference line. Mark off the smaller circle in steps of 30, using OB as the reference line. Set off the axes of the graph alongside as in the previous example. Plot the sine wave of voltage A as before. Plot the sine wave of voltage B in exactly the same way, projecting the first point from B to the y-axis YOY and from each succeeding 30 point to the appropriate 30 point on the horizontal axis of the graph. Points on the resultant graph are plotted by combining the ordinates of A and B at each 30 point. If the graphs lie on the same side of the x-axis, the ordinates are added. If the graphs lie on opposite sides of the axis, the smaller is subtracted from the larger (measurements upwards from the x-axis are positive, measurements downwards are negative). 45

54 Electrical Installation Calculations: Advanced The resultant curve is shown by the dotted line in Figure 42 and its maximum value is approximately 150 V. Its r.m.s. value is = 106 V Example 3 A current of 15 A flows from the 230 V mains at a power factor of 0.76 lagging. Assuming that both current and voltage are sinusoidal, plot graphs to represent them over one cycle. Plot also on the same axes a graph showing the variation in power supplied over one cycle. The procedure for plotting the current and voltage sine waves is the same as that adopted in the previous example. The phase angle between current and voltage is found from the power factor as follows: power factor = cos where is the angle of phase difference cos = 0.76 = U max = = V I max = = A Scales of 20 mm = 200 V and 20 mm = 20 A will be suitable. To obtain the graph of the power supplied, the ordinates of current and voltage are multiplied together (Figure 43). It is convenient to do this every 30 as before. Remember the rules for multiplying positive and negative numbers. Where the resulting graph is negative, additional points are helpful in obtaining a smooth curve. That portion of the power curve lying above the x-axis represents the power supplied to the circuit. That portion lying below the x-axis represents the power returned to the mains from the circuit. 46

55 a.c. waveform and phasor representation + V P = U I U O f = I I Figure 43 Exercise 6 1. Plot a sine wave, over one complete cycle, of an alternating voltage having a maximum value of 325 V. Determine the r.m.s. value of this voltage. 2. An alternating current has the following value taken at intervals of 30 over one half cycle: Angle Current (A) Determine the average and r.m.s. values of this current. 3. Plot a graph over one complete cycle of a sinusoidal alternating voltage having an r.m.s. value of 200 V. 4. Two sinusoidal voltages act in a circuit. Their r.m.s. values are 110 V and 80 V and they are out of phase by 75, the lower voltage lagging. Plot sine waves on the same axes to represent these voltages. Plot a graph of the resultant voltage by adding together the ordinates of the two waves. Give the r.m.s. value of the resultant voltage and state approximately the phase angle between this resultant and the lower voltage. 5. Two alternating currents are led into the same conductor. They are sinusoidal and have r.m.s. values of 4 A and 1 A. The smaller current leads by 120. Plot out the sine waves of these two currents and add the ordinates to obtain the sine wave of the resultant current. Calculate the r.m.s. value of the resultant. 6. The current taken by an immersion heater from the 250 V a.c. mains is 12.5 A. Current and voltage are in phase and are sinusoidal. Plot graphs on the same axes to show the variations in current and voltage over one complete cycle. 47

56 Electrical Installation Calculations: Advanced 7. A 10 F capacitor is connected to a 240 V supply at 50 Hz. The current leads the voltage by 90, and both may be assumed to be sinusoidal. Plot the sine waves of the current and voltage over one complete cycle. 8. A fluorescent lamp takes a current of 1.2 A from a 230 V supply at a power factor of Assuming that both current and voltage are sinusoidal, plot graphs to show how they vary over a complete cycle. 9. The current in a circuit is 25 A and the supply voltage is 220 V. The power factor is 0.6 lagging. Plot sine waves to represent current and voltage over one cycle. Multiply the instantaneous values of current and voltage together to obtain a graph representing the power in the circuit. 10. An inductor of 0.1 H is connected to a 100 V supply at 50 Hz. Neglecting the resistance of the winding, calculate the current which flows. Plot sine waves to represent the current and voltage, assuming that the voltage leads the current by 90. Multiply the ordinates of the waves together to obtain a graph representing the power supplied to the circuit. Phasors Conditions is alternating-current circuits can be represented by means of phasor diagrams. In Figure 44, U is a voltage and I is a current, is the angle of phase difference, and cos is the power factor. Example 1 The current in a circuit is 5 A, the supply voltage is 230 V, and the power factor is 0.8 lagging. Represent these conditions by means of a phasor diagram drawn to scale. f U Lead Lead I I Lag f Lag U (a) Lagging power factor Figure 44 (b) Leading power factor 48

57 a.c. waveform and phasor representation O f = U = 230 V Figure 45 I = 5 A R C U R = 150 V Figure 46 U ~ U C = 200 V Choose a suitable scale (see Figure 45). Power factor = 0.8 = cos cos = 0.8 = Normally the r.m.s. values are used when drawing phasor diagrams. Note that the most accurate construction is obtained by setting off two lines at the required angle and then marking the lines to the appropriate lengths from the point of intersection with compasses which have been set to the respective measurement. Example 2 A resistor and a capacitor are wired in series to an a.c. supply (Figure 46). When a voltmeter is connected across the resistor it reads 150 V. When it is connected to the capacitor terminals it indicates 200 V. Draw the phasor diagram for this circuit to scale and thus determine the supply voltage. As the value of current is not given, it will not be possible to draw its phasor to scale. The current is the same throughout a series circuit and so the current phasor is used as a reference. Draw OI any length to represent the current (Figure 47). 49

58 Electrical Installation Calculations: Advanced O f A U R = 150V I B U C = 200V U = 250V C Figure 47 R L U R U L Figure 48 U = 230 V ~ Equivalent circuit diagram From point O, draw thin lines parallel to and at right angles to OI (capacitor voltage lags behind the current). Choose a suitable scale and use compasses set to the required measurement to mark off OA = U R, the resistor voltage drop in phase with the current and OB = U C, the capacitor voltage drop. With centre A and compasses set to length OB, strike an arc. With centre B and compasses set to OA, strike another arc. These arcs intersect at point C. OC is the resultant voltage, which is equal to the supply voltage. By measurement of OC, the supply voltage is found to be 250 V. Example 3 An inductor takes a current of 5 A from a 230 V supply at a power factor of 0.4 lagging. Construct the phasor diagram accurately to scale and estimate from the diagram the resistance and reactance of the coil. As already explained, although resistance and reactance cannot be separated, it is convenient to draw them apart in an equivalent circuit diagram (Figure 48). The total voltage drop in this case the supply voltage will then be seen to be made up of a resistance voltage drop and a reactance voltage drop. 50

59 a.c. waveform and phasor representation Y P R B U L C Q O U f = U R A I Figure 49 Since, again, we are considering a series circuit in which the current is the same throughout, it is not necessary to draw the current phasor to scale. Power factor = cos where is the angle of phase difference between current and supply voltage and cos = 0.4 = Draw OI any length to represent the current (Figure 49). Choose a suitable scale and set off OC at from OI and of length to represent the supply voltage. Draw OY at right angles to the current phasor and from C draw perpendiculars to cut the current phasor at A and OY at B. The perpendiculars are constructed as follows: (i) Set the compasses to any radius and with centre C draw arcs which cut OY at P and Q. (ii) With the compasses again set to any radius and with centres P and Q strike two more arcs to cut in R. CR is then perpendicular to OY. A similar method is employed in drawing CA. 51

60 Electrical Installation Calculations: Advanced Appliance I = 32 A C I R I C = 8.9 A By measurement, Figure 50 U 250V ~ U R = 93 V U L = 209 V Now U R = I R 93 = 5 R R = 93 5 and = 18.6 U L = I X L (X L is the inductive reactance) 209 = 5 X L X L = = 41.8 Example 4 An appliance takes a single-phase current of 32 A at 0.6 p.f. lagging from a 250 V a.c. supply. A capacitor which takes 8.9 A is wired in parallel with this appliance (Figure 50). Determine graphically the total supply current. As this is a parallel circuit, the voltage is common to both branches and is thus used as the reference phasor. It need not be drawn to scale. Choose a suitable scale. p.f. = cos = 0.6 = 53 8 Draw the voltage phasor (Figure 51) and set off the appliance-current phasor at 53 8 lagging (OA). The capacitor current, 8.9 A, leads on the voltage by 90 and is drawn next (OB). 52